Question

There are m points on a straight line AB and n points on another straight line AC, none of them being A. How many triangles can be formed with these points as vertices?
(a) \[\dfrac{mn}{2}\left( m+n \right)\]
(b) \[\dfrac{mn}{2}\left( m+n-2 \right)\]
(c) \[\dfrac{mn}{2}\left( m+n-1 \right)\]
(d) None of these

Answer 1

Hint: Consider two cases of counting triangles, one in which vertex A is included and the other one in which A is not included. Count all the possible ways of choosing other vertices. Sum up all the possible ways of choosing vertices to get the total number of ways in which triangles can be formed. Use the fact that ‘r’ objects can be chosen from a set of ‘n’ objects in \[{}^{n}{{C}_{r}}\] ways.

Complete step-by-step answer:
We have two straight lines AB and AC such that AB has m points and AC has n points on it other than A. We have to count the number of triangles that can be formed with these vertices.

We will consider two cases to count the triangles – one in which vertex A is included and the other one in which A is not included. We will count all the possible ways to choose vertices in each of the cases and then sum up the values to get the total number of triangles that can be formed.
We know that ‘r’ objects can be chosen from a set of ‘n’ objects in \[{}^{n}{{C}_{r}}\] ways.

Case 1: Vertex A is included in the triangle.
We already have 1 vertex. We have two choose two other vertices.
We know that more than two vertices on the same line can’t form a triangle. So, we must choose one vertex from each of the lines.
Thus, the number of ways to choose one vertex from m points \[={}^{m}{{C}_{1}}=\dfrac{m!}{1!\left( m-1 \right)!}=m\].
Similarly, the number of ways to choose one vertex from n points \[={}^{n}{{C}_{1}}=\dfrac{n!}{1!\left( n-1 \right)!}=n\].
Thus, the total number of ways to choose two vertices \[={}^{m}{{C}_{1}}\times {}^{n}{{C}_{1}}=mn\].
Hence, we can form ‘mn’ triangles such that vertex A is included in the triangle.

Case 2: Vertex A is not included in the triangle.
We have to choose all the three vertices. So, we must choose two vertices from one line and one from the other one.
If we choose two vertices from line AB and one from AC, we need to choose 2 points out of m points and 1 point out of n points.
Number of ways to choose 2 vertices from m vertices \[={}^{m}{{C}_{2}}=\dfrac{m!}{2!\left( m-2 \right)!}=\dfrac{m\left( m-1 \right)}{2}\].
Similarly, the number of ways to choose one vertex from n points \[={}^{n}{{C}_{1}}=\dfrac{n!}{1!\left( n-1 \right)!}=n\].
Total number of ways to choose 2 vertices from AB and 1 from AC \[={}^{m}{{C}_{2}}\times {}^{n}{{C}_{1}}=\dfrac{mn\left( m-1 \right)}{2}\].
Now, we will calculate the number of ways to choose two points from AC and one from AB.
Number of ways to choose 2 vertices from n vertices \[={}^{n}{{C}_{2}}=\dfrac{n!}{2!\left( n-2 \right)!}=\dfrac{n\left( n-1 \right)}{2}\].
Similarly, the number of ways to choose one vertex from m points \[={}^{m}{{C}_{1}}=\dfrac{m!}{1!\left( m-1 \right)!}=m\].
Total number of ways to choose 2 vertices from AC and 1 from AB \[={}^{m}{{C}_{1}}\times {}^{n}{{C}_{2}}=\dfrac{mn\left( n-1 \right)}{2}\].
Total number of ways to choose 3 vertices is the sum of the number of ways of choosing 2 vertices from AB and 1 from AC with the number of ways of choosing 2 vertices from AC and 1 from AB.
Thus, total number of ways to choose 3 vertices \[=\dfrac{mn\left( n-1 \right)}{2}+\dfrac{mn\left( m-1 \right)}{2}=\dfrac{mn\left( m+n-2 \right)}{2}\].
Total number of triangles that can be formed \[=mn+\dfrac{mn\left( m+n-2 \right)}{2}=\dfrac{mn\left( 2+m+n-2 \right)}{2}=\dfrac{mn\left( m+n \right)}{2}\].

Hence, the number of triangles that can be formed is \[\dfrac{mn\left( m+n \right)}{2}\], which is option (a).

Note: It’s necessary to consider each of the two cases separately. Also, we will get an incorrect answer if we don’t consider the fact that three points on the same line can’t form a triangle as they are collinear. We must know that \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].